mathmath333
  • mathmath333
Logarithm question
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align}& \text{solve}\hspace{.33em}\\~\\& \log_{2}(9-2^{x})=10^{\log_{10}(3-x)}\hspace{.33em}\\~\\ \end{align}}\)
mathmath333
  • mathmath333
@freegirl11112
mathmath333
  • mathmath333
@freckles

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anonymous
  • anonymous
hi I dot know this yet sorry
mathmath333
  • mathmath333
mistyped ur name was calling frekels
freckles
  • freckles
\[\log_2(9-2^x)=10^{\log(3-x)} \\ \log_2(9-2^x)=3-x \\ 2^{3-x}=9-2^x \\ 2^{3-x}+2^x=9 \\ 2^3 2^{-x}+2^x=9 \\ 8 \cdot 2^{-x}+2^x=9 \\ \text{ multiply both sides by } 2^x \\ 8 +2^{2x}=9 \cdot 2^{x}\]
mathmath333
  • mathmath333
i didn't understand the second step \(\log_2(9-2^x)=3-x\)
Nnesha
  • Nnesha
what happened to 10^log ?? :o
freckles
  • freckles
sorry log_10(x) and 10^x are inverses
freckles
  • freckles
\[10^{\log_{10}(3-x)}=3-x\] when 3-x>0
Nnesha
  • Nnesha
mhm oh okay thanks:=)
freckles
  • freckles
and I'm so sorry but I have to go someone is waiting on me
freckles
  • freckles
hey @mathmath333 were you able to finish
freckles
  • freckles
you have a quadratic in terms of 2^x
mathmath333
  • mathmath333
yep o got it.

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